Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $z = \dfrac{-2k + 2}{k^3 + 7k^2 - 8k} \times \dfrac{3k^3 - 9k^2 - 30k}{3k - 15} $
Solution: First factor out any common factors. $z = \dfrac{-2(k - 1)}{k(k^2 + 7k - 8)} \times \dfrac{3k(k^2 - 3k - 10)}{3(k - 5)} $ Then factor the quadratic expressions. $z = \dfrac {-2(k - 1)} {k(k - 1)(k + 8)} \times \dfrac {3k(k - 5)(k + 2)} {3(k - 5)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {-2(k - 1) \times 3k(k - 5)(k + 2) } { k(k - 1)(k + 8) \times 3(k - 5)} $ $z = \dfrac {-6k(k - 5)(k + 2)(k - 1)} {3k(k - 1)(k + 8)(k - 5)} $ Notice that $(k - 1)$ and $(k - 5)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {-6k(k - 5)(k + 2)\cancel{(k - 1)}} {3k\cancel{(k - 1)}(k + 8)(k - 5)} $ We are dividing by $k - 1$ , so $k - 1 \neq 0$ Therefore, $k \neq 1$ $z = \dfrac {-6k\cancel{(k - 5)}(k + 2)\cancel{(k - 1)}} {3k\cancel{(k - 1)}(k + 8)\cancel{(k - 5)}} $ We are dividing by $k - 5$ , so $k - 5 \neq 0$ Therefore, $k \neq 5$ $z = \dfrac {-6k(k + 2)} {3k(k + 8)} $ $ z = \dfrac{-2(k + 2)}{k + 8}; k \neq 1; k \neq 5 $